Monday, August 10, 2020

There are two grandparents, two parents and three grandchildren in a family. The average age of the grandparents is 67 years, that of the parents is 35 years and that of the grandchildren is 6 years. What is the average age of the family?

A.  \(33\frac{2}{6}\)
B.  \(36\frac{8}{3}\)
C.  \(34\frac{4}{2}\)
D.  \(31\frac{5}{7}\)

Answer:

D.  \(31\frac{5}{7}\)

Solution:

\[\left. \begin{array}{l}\\ { \text {Sum of the Ages of Grand Parents:}}\\ { \text {Avg Age=}}\frac{\text {Sum of Ages}}{\text {Total Persons}}\\ 67=\frac{\text {Sum of Ages}}{2}\\ \text {Sum of Ages}=67\times2=134 …(a)\\ { \text {Sum of the Ages of Parents:}}\\ =35\times2=70 … (b)\\ { \text {Sum of the Ages of Children:}}\\ =6\times3=18 ….(c)\\ { \text {Total Family Members:}}\\ =2+2+3=7 ….(d)\\ { \text {Average Age of family=}} (\frac{\text {Sum of the Ages}}{\text {Total Persons}}) \\ { \text {Average Age of family=}} (\frac{a+b+c}{d}) \\ { = ( \frac { 134 + 70 + 18 } { 7 } ) }\\{ = \frac { 222 } { 7 } }\\{ = 31 \frac { 5 } { 7 } \text { years. } }\end{array} \right.\]

A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 6500 for the last 6 months?

A. Rs. 5081
B. RS. 4810
C. RS. 4991
D. Rs. 4972

Solution:

\begin{array} { l } \bf \text { Data we Have: }\\ \text { Required Avg Sale for 6 month= 6500 }\\ \text { Total Number of months= 6 }\\ { \text { Total sale for } 5 \text { months =}} \\ \Rightarrow{ R s . ( 6435 + 6927 + 6855 + 7230 + 6562 )}\\ { \Rightarrow \text { Rs. } 34009 \text { } } \\ \text { Formula for the required sale will be:}\\ \text { Required Sale = }\frac{\text { Sum of Sale of 6 months }}{\text { Total number of Months }}\\ \\ \bf \text { By Putting the values we get:}\\ 6500= \frac{\text { Sum of Sale of 6 months }}{6}\\ \text { Sum of Sale of 6 months= }6500\times6=39000\\ { \therefore \text { Required sale } = ( 39000 -\text { 5 month sale )} } \\ { = R s . ( 39000 – 34009 ) } \\ { = \text { Rs. } 4991 } \end{array}

The average weight of 8 person’s increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What might be the weight of the new person?

A. 84 kg
B. 81 kg
C. 80 kg
D. 85 kg

Solution:

 

\[\left. \begin{array} { l }\text {Average weight of each person = 65kg} \\ \text {Avg weight of each person increased = }2.5kg\\ \text {Total avg weight of 8 persons increased by:} \\ 8\times2.5=20kg\\ \text {Since total weight increased is actually} \\ \text {the weight of new person. Therefore,}\\ \text { The weight of new person = 65+20= 85KG}\\ \end{array} \right.\]

The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. What is the average age of the team?

A. 20 years
B. 27 years
C. 21 years
D. 23 years

Solution:

\[\left. \begin{array} { l } \text {Captain Age = 26 years}\\ \text {Wicket Keepers age = 26+3= 29 years}\\ \text {Let the average age of team be x years}\\ \text {So, avg age of 9 players would be = 9(x-1)}\\ \text {Total number of players = 11}\\ \text {Putting the values in the formula we get:}\\ \text {Avg age of team =}\frac{\text {Sum of the Ages of Players}}{\text {Total number of Players}}\\ \text {x = } \frac{26+29+9(x-1)}{11}\\ \text {11 = 55 + 9x – 9}\\ \text {11x – 9x = 46}\\ \text {2x = 46} \text {x = 23 years} \end{array} \right.\]

The average monthly income of P and Q is Rs. 5050. The average monthly income of Q and R is Rs. 6250 and the average monthly income of P and R is Rs. 5200. The monthly income of P is:

\[\left. \begin{array} { l }\text {Avg monthly income of P and Q =5050 }\\ \text {Sum of incomes of P and Q is:}\\ \text {Avg income =} \frac{\text {Sum of Incomes}}{\text {Total Number of persons}}\\ \Rightarrow \text {5050 =} \frac{P+Q}{2}\\ \text {P+Q = 5050}\times 2\text {= 10100 …(i)}\\ \text {Likewise:} \text {Q+R = 6250}\times 2\text {= 12500 …(ii)}\\ \text {P+R = 5200}\times 2\text {= 10400 …(iii)}\\ { \text { Now adding (i), (ii) and (iii), we get: }}\\ \Rightarrow { 2 ( P + Q + R ) = 33000}\\ \text { P + Q + R= 16500 …(iv) } \\ { \text { Subtracting (ii) from (iv), we get } P = 4000 } \\ { \therefore \text { P’s monthly income } = \text { Rs. } 4000 } \end{array} \right.\]

A car owner buys petrol at Rs.7.50, Rs. 8 and Rs. 8.50 per litre for three successive years. What approximately is the average cost per litre of petrol if he spends Rs. 4000 each year?

A. 8.71
B. 7.98
C. 6.87
D. 9.99

 

\[\left. \begin{array} { l } \text {Total expenditure on petrol each year = 4000}\\ \text {Petrol consumption in 1st year = }\frac{4000}{7.5}\\ \text {Petrol consumption in 2nd year = }\frac{4000}{8}\\ \text {Petrol consumption in 3rd year = }\frac{4000}{8.50}\\ \text {Total quantity of petrol consumed in 3 years:}\\ \Rightarrow ( \frac { 4000 } { 7.50 } + \frac { 4000 } { 8 } + \frac { 4000 } { 8.50 } ) \text { litres } \\ { = 4000 ( \frac { 2 } { 15 } + \frac { 1 } { 8 } + \frac { 2 } { 17 } ) \text { litres } }\\ {= ( \frac { 76700 } { 51 } ) \text { litres } }\\ { \text { Total amount spent } = \text { RS. } ( 3 \times 4000 )}\\ {= \text { Rs. } 12000 \text { . } } \\ { \therefore \text { Average cost } = \text { Rs. } ( \frac { 12000 \times 51 } { 76700 } ) }\\ {\text {= Rs. } \frac { 6120 } { 767 } = Rs 7 .98 } \end{array} \right.\]

The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, then the weight of B is:

A. 29 kg
B. 31 kg
C. 40 kg
D. 12 kg

Solution:

Let A, B, C represent their respective weights. Then, we have:
\[\left. \begin{array} { l } { A + B + C = ( 45 \times 3 ) = 135 \ldots . \text { (i) } } \\ { A + B = ( 40 \times 2 ) = 80 \ldots . \text { (ii) } } \\ { B + C = ( 43 \times 2 ) = 86 \ldots \text { (iii) } } \\ { \text { Adding (ii) and (iii), we get: } A + 2 B + C }\\
{= 166 \ldots \text { (iv) } } \\ { \text { Subtracting (i) from (iv), we get: } B = 31 . } \\ { \therefore \text { B’s weight } = 31 kg \text { . } } \end{array} \right.\]

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